I'm just your average idiot, not a rocket programmer or electrical engineer.
If this blows up your model, I will feel bad.
But I won't buy you another one. That said, here goes!

So.... I said I would do it, and here it is.
A single 2N2222 transistor switching 4 LED's while draining less
than 1 milliamp from the host LED circuit.
The drain on the host is so minute, it should not effect even the most
sensitive equipment.

I emulated a 3S pack with my power supply, (12.6V) so resistor values will
need to be tailored to your application.

The breadboard (or proto-board) is a Radio Shack part. The internal connection
layout can be found at the bottom of this page here:

Actual resistor values will also depend on what voltage potential with respect to ground
is available from the donor LED output.

The red wire coming out of the breadboard is the gate with a 10K
ohm resistor in series. This provides 10K of isolation from the host.
Good enough for what we are doing. We could use opto-isolators
and all kinds of fancy stuff, but we are dealing with aircraft- lighter
is better.

*EDIT - I just learned that opto-isolators are a stand-alone device, and
are probably the safest solution. They also weigh little more than a
single transistor.

*CONTINUED

I put a 1N4004 diode in series with the input (gate) to
protect our LED source circuit, so that no current will flow from
our light project to the circuit we are tapping in to (that was added later,
and is not shown in all the pictures). Any 4000 series diode should work,
I picked a 1N4004 because I have lots of them.

The resistors in series with the output LED's (the ones in a row)
are all 510 ohm, and if all your LED's are the same you could substitute
a single resistor.
But we like many colors, and different color LED's require different voltages
to attain the same brightness levels. Also, some LED's are much more
power hungry than others (some red LED's need 10mA... others need 50).
LED current needs to be adjustable individually, hence the individual resistors.
Your actual values will be dictated by LED type, and your supply voltage.

The single 100 ohm resistor is there so we don't set our model on fire if
there is a short somewhere. Basically current limiting.

This is so simple, you don't even need a PCB.

The measured drain from the host in this example was 410 microamps.
(0.41 milliamps, or .00041 amps) as shown on the Fluke.
Total draw for the entire circuit is 44mA, (shown on the power supply)
but that will vary with different LED's.

When I supply power (less than 1mA) to the red wire (the gate) the LED's
turn on. When voltage is removed from the gate, they all go out

Have fun Mike !!

UPDATE:

Tried this on my 9116 and found that you have to omit the diode.
The extra voltage drop is just too much. The 10K resistor keeps
any significant amount of current from flowing anyways.
I tapped into the flashing LED on the PCB (it flashes when the radio
is off) and it worked perfectly, flashing all 4 LED's and drawing almost
no current from the PCB.
The gate drew so little, it was beyond what my Fluke could measure, and
bounced around at .01mA.

These were fairly efficiency LED's, and only need 10mA each.
Some LED's draw much more, and you may only be able to
drive one or two of them before the transistor gets warm.
Average current consumption will be listed on the LED datasheet.

So.... I said I would do it, and here it is.
A single 2N2222 transistor switching 4 LED's while draining less
than 1 milliamp from the host LED circuit.
The drain on the host is so minute, it should not effect even the most
sensitive equipment.

I emulated a 3S pack with my power supply, (12.6V) so resistor values will
need to be tailored to your application.

Actual values will also depend on what voltage potential with respect to ground
is available from the donor circuit.

The red wire coming out of the breadboard is the gate with a 10K
ohm resistor in series. This provides 10K of isolation from the host.
Good enough for what we are doing. We could use opto-isolators
and all kinds of fancy stuff, but we are dealing with aircraft- lighter
is better.

I put a 1N4004 diode in series with the input (gate) to
protect our LED source circuit, so that no current will flow from
our light project to the circuit we are tapping in to (that was added later,
and is not shown in all the pictures). Any 4000 series diode should work,
I picked a 1N4004 because I have lots of them.

The resistors in series with the output LED's (the ones in a row)
are all 510 ohm, and if all your LED's are the same you could substitute
a single resistor.
But we like many colors, and different color LED's require different voltages
to attain the same brightness levels. Also, some LED's are much more
power hungry than others (some red LED's need 10mA... others need 50).
LED current needs to be adjustable individually, hence the individual resistors.
Your actual values will be dictated by LED type, and your supply voltage.

The single 100 ohm resistor is there so we don't set our model on fire if
there is a short somewhere. Basically current limiting.

This is so simple, you don't even need a PCB.

The measured drain from the host in this example was 410 microamps.
(0.41 milliamps, or .00041 amps) as shown on the Fluke.
Total draw for the entire circuit is 44mA, (shown on the power supply)
but that will vary with different LED's.

When I supply power (less than 1mA) to the red wire (the gate) the LED's
turn on. When voltage is removed from the gate, they all go out

Have fun Mike !!

UPDATE:

I tried powering this with my Double Horse, and found that the
resistor values on the gate input need to vary by quite a large
margin. When I tapped into the flashing LED on my DH9116
PCB (it flashes when the radio is off) I had to use a 1K ohm
resistor on the gate in place of the 10K, so enough current would
flow to switch on the transistor and 4 LED's. Keep in mind, the
9116 runs on 8.4V, and all my values were originally for a 3S pack
that puts out 12.6V.

This will vary from model to model. Start big, and go smaller,
until you get enough juice to make the transistor conduct.

Diodes are such a common component, it is a good investment
if only to know you have some if you need them.

4000 series cost about a penny in bulk, hope you didn't spend too much

Also, these resistor values are not set in stone, I would recommend an assortment
of 1/4 watt. I used a 1/2 watt for the current limiter, but a 1/4 watt would have been
fine in this case. The 10K gate can use the smallest wattage resistor you can solder.

Another factor is how the donor switches the LED. Some use a switched
ground. Most are probably always grounded, and switch the positive side.
In rare cases I'm thinking you might need a PNP transistor. We will take it
as it comes.

yup what you have there is a really simple transistor switch. what is funny is, before I even read it, I said to myself. "a simple 2n2222 would do it" lol

you dont have to use a seperate resistor for each LED either, unless they are different brightness.

as you said you could use opto-isolators. the little dip4 packaged ones. they are even easier than that circuit.

they have "switch" input, which is jsut an LED internally and then a fet output. you ground the cathode and put a 1k ohm resistor in series with the input of the isolator and the output of the driver circuit.

for the ouput of the isolator, you put B+ on one side of the fet and a 1k resistor (or whatever value you want to make the LEDs bright enough) with the LEDs. The driver circuit will see 1 LED at about 10mA of current or so and your opto will switch about 500mA - 1A depending on the size of it. you dont have to worry about forward bias of the BJT or diode drops either.

Only question I would have would be if the opto-isolator has an actual LED in it,
wouldn't it draw the same as an LED?

What we wanted to do was not draw any extra current from the host circuit.

For example, lets say he wants to keep the "donor" LED in place, as well as drive
other LED's (which I believe he does).

Then you would have the 10mA or so that would normally be drawn by the LED,
AND the current draw for the opto-isolator- effectively doubling the current draw
being pulled from the donor circuit. (20mA instead of 10mA)
I was worried about how this might effect parameters of the donor LED
driver circuit.

I believe it is driven from the RX "CPU" because it will flash when battery
voltage gets low. (the whole point of this is to make LED's on the OUTSIDE
of the model flash when batteries get low, instead of one tucked away
inside the model.

If it is running directly off an output from the processor,
it might have very low current capabilities.

I suppose this might be ok (drawing 20mA instead of 10), but that's why I
wanted the opinion of someone who knows what the heck they are doing.

With the transistor, I wasn't even pulling 1mA from the donor circuit.
It would never even know we were tapped in, as the load on the "donor"
circuit would change by less than 10%, an acceptable tolerance in most circuits.

But in my ignorance was unsure if biasing and isolation were correct.

Come to think of it, I see no reason to keep the donor LED in place if it
is buried in the model. The opto-isolator in place of the original LED looks
like the way to go after all. Current draw would be the same, and we
would not have to worry about biasing or isolation.

Thanks for taking the time to check this out.
I know how it is, doing work related stuff in your free time (it sucks)

What we wanted to do was not draw any extra current from the host circuit.
For example, lets say he wants to keep the "donor" LED in place, as well as drive
other LED's (which I believe he does).

Then you would have the 10mA or so that would normally be drawn by the LED,
AND the current draw for the opto-isolator- effectively doubling the current draw
being pulled from the donor circuit. (20mA instead of 10mA)
I was worried about how this might effect parameters of the donor LED
driver circuit.

I believe it is driven from the RX "CPU" because it will flash when battery
voltage gets low. (the whole point of this is to make LED's on the OUTSIDE
of the model flash when batteries get low, instead of one tucked away
inside the model)

If it is running directly off an output from the processor,
it might have very low current capabilities.

I suppose this might be ok (drawing 20mA instead of 10), but that's why I
wanted the opinion of someone who knows what the heck they are doing.

With the transistor, I wasn't even pulling 1mA from the donor circuit.
It would never even know we were tapped in, as the load on the "donor"
circuit would change by less than 10%, an acceptable tolerance in most circuits.

But in my ignorance was unsure if biasing and isolation were correct.

Yes, Steve,
my intentions are to use the onboard SMD-Leds to trigger the much more powerful Nav-LEDS without influencing the really small 4-in-one PCB of my X4s. So it's really importantfor me to draw as less current as possible from the two LED-sockets - just to play it safe. I will not get any replacement PCB in near future for these quads - they will not be produced for months as a spare ....
-mike-